I am working on cross word algorithm to develop a word app. After doing a lot of googling or search on StackOverflow, I was able to reach this point. But yet I am not able to understand the right implementation for algorithm in Java. Below is the class I used.
public class Crosswords {
char[][] cross;
int rows;
int cols;
char[][] numberGrid;
boolean startword;
final char DEFAULT = ' ';
public Crosswords() {
rows = 50;
cols = 50;
cross = new char[rows][cols];
numberGrid = new char [rows][cols];
for (int i = 0; i < cross.length;i++){
for (int j = 0; j < cross[i].length;j++){
cross[i][j] = DEFAULT;
}
}
}
public Crosswords(int ros, int colls) {
rows = ros;
cols = colls;
cross = new char[rows][cols];
numberGrid = new char [rows][cols];
for (int i = 0;i < cross.length; i++){
for (int j = 0; j < cross[i].length; j++){
cross[i][j] = DEFAULT;
}
}
}
public String toString() {
String s = new String();
//String d = new String();
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++){
s = s + cross[i][j] + " ";
}
s = s + "\n";
}
return s;
}
public void addWordh(String s, int r, int c) {
int i = 0;
int j = 0;
boolean b = true;
boolean intersectsWord = true;
if (s.length() > cols) {
System.out.println(s + " is longer than the grid. Please try another word.");
return;
}
if (c + s.length() > cols) {
System.out.println(s + " is too long. Please try another word.");
return;
}
if ((r - 2) >= 0) {
if ((cross[r - 1][c - 1 + s.length()] == DEFAULT) || (cross[r - 1][c - 1 + s.length()] == '*')) {
intersectsWord = false;
}
else { intersectsWord = true;}
if (intersectsWord == true) {
System.out.println("The word " + s + " intersects the beginning of another word!");
return;
}
}
for (i = 0; i < s.length(); i++) {
if ((cross[r - 1][c - 1 + i] == DEFAULT) || (cross[r - 1][c - 1 + i] == s.charAt(i))) {
b = true;
}
else {
b = false;
System.out.println("Unable to add " + s + ". Please try another word.");
return;}
}
if (b == true) {
if ((s.length() <= cols) && (c + s.length() <= cols) &&
(cross[r - 1][c - 1] == s.charAt(0)) || (cross[r - 1][c - 1] == DEFAULT)) {
while (j < s.length()) {
cross[r - 1][c - 1 + j] = s.charAt(j);
if (j==0){
startword = true;
}
cross[rows - 1 - (r - 1)][cols - 1 - (c - 1 + j)] = '*';
j++;
}
}
}
}
public void addWordv(String s, int r, int c) {
int i = 0;
int j = 0;
boolean b = true;
boolean intersectsWord = true;
if (s.length() > rows) {
System.out.println(s + " is longer than the grid. Please try another word.");
}
if (r + s.length() > rows) {
System.out.println(s + " is too long. Please try another word.");
}
else {
if ((r - 2) >= 0) {
if ((cross[r - 2][c - 1] == DEFAULT) || (cross[r - 2][c - 1] == '*')) {
intersectsWord = false;
}
else { intersectsWord = true;}
if (intersectsWord == true) {
System.out.println("The word " + s + " intersects the end of another word!");
return;
}
}
if ((cross[r - 1 + s.length()][c - 1] == DEFAULT) || (cross[r - 1 + s.length()][c - 1] == '*')) {
intersectsWord = false;
}
else { intersectsWord = true;}
if (intersectsWord == true) {
System.out.println("The word " + s + " intersects the end of another word!");
return;
}
for (i = 0; i < s.length(); i++) {
if ((cross[r - 1 + i][c - 1] == DEFAULT) || (cross[r - 1 + i][c - 1] == s.charAt(i))) {
b = true;
}
else {
b = false;
System.out.println("Unable to add " + s + ". Please try another word.");
return;}
}
if (b == true) {
if ((s.length() <= rows) && (r + s.length() <= cols) &&
(cross[r - 1][c - 1] == s.charAt(0)) || (cross[r - 1][c - 1] == DEFAULT)) {
while (j < s.length()) {
cross[r - 1 + j][c - 1] = s.charAt(j);
if (j==0){
startword = true;
}
cross[rows - 1 - (r - 1 + j)][cols - 1 - (c - 1)] = '*';
j++;
}
}
}
}
}
public void setNumberGrid(){
numberGrid = new char [rows][cols];
for (int i = 0; i < cross.length; i++){
for (int j=0; j < cross[rows].length; j++){
if (cross[i][j] == DEFAULT){
numberGrid[i][j] = (char) 0;
}
else if (startword == true){
numberGrid[i][j] = (char) -2;
}
else {
numberGrid[i][j] = (char) -1;
}
}
int count = 1;
for (i=0; i < cross.length; i++){
for (int j=0; j < cross[rows].length; j++){
if (numberGrid[i][j] == -2){
numberGrid[i][j] = (char)count;
count++;
}
}
}
}
}
public String printNumberGrid() {
for (int i=0; i < cross.length; i++){
for (int j=0; j < cross[rows].length; j++){
if (numberGrid[i][j] == (char)-1){
numberGrid[i][j] = ' ';
}
else if (numberGrid[i][j] == (char)0){
numberGrid[i][j] = '#';
}
}
}
String d = new String();
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++){
d = d + numberGrid[i][j] + " ";
}
d = d + "\n";
}
return d;
}
public static void main(String[] args) {
Crosswords g = new Crosswords();
g.addWordv("rawr", 4, 5);
g.addWordh("bot", 5, 4);
g.addWordv("raw", 7, 5);
g.addWordh("cat", 4, 5);
g.addWordh("bass", 6, 10);
System.out.println(g);
Crosswords c = new Crosswords(20, 20);
c.addWordh("HELLO", 1, 1);
c.addWordv("HAPLOID", 1, 1);
c.addWordh("COMPUTER", 3, 12);
c.addWordv("CAT", 2, 11);
c.addWordv("WOAH", 2, 20);
c.addWordh("PARKING", 20, 5);
c.addWordv("ARK", 17, 6);
c.addWordh("AHOY", 6, 18);
c.addWordv("AHOY", 18, 10);
c.addWordv("ADVANTAGE", 2, 12);
c.addWordv("INTERNAL", 2, 18);
c.addWordh("BANTER", 7, 11);
c.addWordv("BEAGLE", 5, 12);
c.addWordh("BASE", 8, 3);
c.addWordv("BALL", 8, 3);
c.addWordh("LEFT", 10, 3);
c.addWordv("SAFE", 8, 5);
System.out.print(c);
}
}
As you can see in Main method that i am adding the words but also giving the row and column number to place the words like c.addWordv("Safe",8,5); where 8 and 5 is column number.
Now Question is how can i implement cross word algorithm which just take words and place them on board randomly without taking the row and column numbers. Thanks in advance
EDIT:
I want to modify this class algo the way that i dont have to give away the rows and columns number..
//Pseudo Code
If the crossword size is maxSize and any word's length is stored in wordLength ,then you can use random method as below int maxSize=20; int wordLength=4;
Random random =new Random();
int r,c;
//for horizontal
r=random.nextInt(maxSize-wordLength);
c=random.nextInt(maxSize);
//for vertical
r=random.nextInt(maxSize);
c=random.nextInt(maxSize-wordLength);
You can store the row and column and generate the new one if its already present.